the partial derivative. Thus, ∂w
∂x
y,z
would signify the partial derivative ofw with respect to x, while holding bothyandzconstant. Hence,x,y, andzare the complete set of independent variables in this case. On the other hand, we would use(∂w/∂x)yto indicate thatxandyalone are the independent variables. In the case thatw=x+2y+z, this notation gives
∂w
∂x
y,z
=1, ∂w
∂y
x,z
=2, and ∂w
∂z
x,y
=1. Ifz=x y, then we also have
∂w
∂x
y
=1+y, and ∂w
∂y
x =2+x. In this way, the ambiguity of notation can be avoided. Use this notation in Exercises 39–45.
39. Letw=x+7y−10zandz=x2+y2. (a) Find
∂w
∂x
y,z
, ∂w
∂y
x,z, ∂w
∂z
x,y
, ∂w
∂x
y
, and
∂w
∂y
x.
(b) Relate (∂w/∂x)y,z and (∂w/∂x)y by using the chain rule.
40. Repeat Exercise 39 wherew=x3+y3+z3andz= 2x−3y.
41. Suppose s=x2y+x zw−z2 and x yw−y3z+x z
=0. Find ∂s
∂z
x,y,w
and ∂s
∂z
x,w.
42. LetU=F(P,V,T) denote the internal energy of a gas. Suppose the gas obeys the ideal gas lawP V =kT, wherekis a constant.
(a) Find ∂U
∂T
P
. (b) Find
∂U
∂T
V
. (c) Find
∂U
∂P
V
.
43. Show that ifx,y,zare related implicitly by an equation of the formF(x,y,z)=0, then
∂x
∂y
z
∂y
∂z
x
∂z
∂x
y
= −1.
This relation is used in thermodynamics. (Hint: Use Exercise 36.)
44. The ideal gas law P V =kT, wherek is a constant, relates the pressureP, temperatureT, and volumeV of a gas. Verify the result of Exercise 43 for the ideal gas law equation.
45. Verify the result of Exercise 43 for the ellipsoid ax2+by2+cz2=d
wherea,b,c, anddare constants.
An alternative way to view ∂f∂x(a,b) is as the rate of change of f as we move
“infinitesimally” froma=(a,b) in thei-direction, as suggested by Figure 2.67.
This is easy to see since, by the definition of the partial derivative,
∂f
∂x(a,b)= lim
h→0
f(a+h,b)− f(a,b) h
= lim
h→0
f((a,b)+(h,0))− f(a,b) h
= lim
h→0
f((a,b)+h(1,0))− f(a,b) h
= lim
h→0
f(a+hi)− f(a)
h .
Note that we are identifying the point (a,b) with the vectora=(a,b)=ai+bj.
Similarly, we have
∂f
∂y(a,b)= lim
h→0
f(a+hj)− f(a)
h .
Writing partial derivatives as we just have enables us to see that they are special cases of a more general type of derivative. Supposevis any unit vector in R2. (The reason for taking a unit vector will be made clear later.) The quantity
h→0lim
f(a+hv)− f(a)
h (1)
is nothing more than the rate of change of f as we move (infinitesimally) froma= (a,b) in the direction specified byv=(A,B)= Ai+Bj. It’s also the slope of the curve obtained as the intersection of the surfacez= f(x,y) with the vertical plane B(x−a)−A(y−b)=0. (See Figure 2.68.) We can use the limit expression in (1) to define the derivative of any scalar-valued function in a particular direction.
i y
x
z
(a, b, f(a, b)) z = f(x, y)
Figure 2.67 Another way to view the partial derivative∂f/∂xat a point.
x y
z
(a, b, f(a, b))
B(x − a) − A(y − b) = 0 z = f(x, y)
v
Figure 2.68 The directional derivative.
DEFINITION 6.1 Let X be open inRn, f:X⊆Rn →R a scalar-valued function, anda∈ X. Ifv∈Rnis any unit vector, then thedirectional deriva- tive of f at a in the direction of v, denotedDvf(a), is
Dvf(a)= lim
h→0
f(a+hv)− f(a) h
(provided that this limit exists).
EXAMPLE 1 Suppose f(x,y)=x2−3x y+2x−5y. Then, ifv=(v, w)∈ R2is any unit vector, it follows that
Dvf(0,0)= lim
h→0
f((0,0)+h(v, w))− f(0,0) h
= lim
h→0
h2v2−3h2vw+2hv−5hw h
= lim
h→0(hv2−3hvw+2v−5w)
=2v−5w.
Thus, the rate of change of f is 2v−5w if we move from the origin in the direction given by v. The rate of change is zero if v=(5/√
29,2/√ 29) or (−5/√
29,−2/√
29). ◆
Consequently, we see that the partial derivatives of a function are just the “tip of the iceberg.” However, it turns out that when f is differentiable, the partial derivatives actually determine the directional derivatives for all directionsv. To see this rather remarkable result, we begin by defining a new function F of a single variable by
F(t)= f(a+tv). Then, by Definition 6.1, we have
Dvf(a)=lim
t→0
f(a+tv)− f(a)
t =lim
t→0
F(t)−F(0)
t−0 =F(0). That is,
Dvf(a)= d
dt f(a+tv)
t=0. (2)
The significance of equation (2) is that, when f is differentiable ata, we can apply the chain rule to the right-hand side. Indeed, letx(t)=a+tv. Then, by the chain rule,
d
dt f(a+tv)= D f(x)Dx(t)= D f(x)v. Evaluation att =0 gives
Dvf(a)= D f(a)v= ∇f(a)·v. (3) The purpose of equation (3) is to emphasize the geometry of the situation. The result above says that the directional derivative is just the dot product of the
gradient and the direction vectorv. Since the gradient is made up of the partial derivatives, we see that the more general notion of the directional derivative depends entirely on just the direction vector and the partial derivatives. To be more formal, we summarize this discussion with a theorem.
THEOREM 6.2 Let X ⊆Rn be open and suppose f:X →R is differentiable at a∈X. Then the directional derivative Dvf(a) exists for all directions (unit vectors)v∈Rnand, moreover, we have
Dvf(a)= ∇f(a)·v.
EXAMPLE 2 The function f(x,y)=x2−3x y+2x−5y we considered in Example 1 has continuous partials and hence, by Theorem 3.5, is differentiable.
Thus, Theorem 6.2 applies to tell us that, for any unit vectorv=vi+wj∈R2, Dvf(0,0)= ∇f(0,0)·v=(fx(0,0)i+ fy(0,0)j)·(vi+wj)
=(2i−5j)·(vi+wj)
=2v−5w,
as seen earlier. ◆
EXAMPLE 3 The converse of Theorem 6.2 does not hold. That is, a function may have directional derivatives in all directions at a point yet fail to be differ- entiable. To see how this can happen, consider the function f:R2→Rdefined by
f(x,y)=
⎧⎪
⎨
⎪⎩ x y2
x2+y4 if (x,y)=(0,0) 0 if (x,y)=(0,0) .
This function is not continuous at the origin. (Why?) So, by Theorem 3.6, it fails to be differentiable there; however, we claim that all directional derivatives exist at the origin. To see this, let the direction vectorvbevi+wj. Hence, by Definition 6.1, we observe that
Dvf(0,0)= lim
h→0
f((0,0)+h(vi+wj))− f(0,0) h
= lim
h→0
1 h
hv(hw)2 (hv)2+(hw)4 −0
= lim
h→0
h2vw2 h2(v2+h2w4)
= lim
h→0
vw2
v2+h2w4 = vw2 v2 = w2
v .
Thus, the directional derivative exists wheneverv=0. Whenv=0 (in which casev=j), we, again, must calculate
Djf(0,0)= lim
h→0
f((0,0)+hj)− f(0,0) h
= lim
h→0
f(0,h)− f(0,0) h
= lim
h→0
0−0 h =0.
Consequently, this directional derivative (which is, in fact,∂f/∂y) exists as well.
◆ The reason we have restricted the direction vectorvto be of unit length in our discussion of directional derivatives has to do with the meaning ofDvf(a), not with any technicalities pertaining to Definition 6.1 or Theorem 6.2. Indeed, we can certainly define the limit in Definition 6.1 for any vectorv, not just one of unit length. So, supposewis an arbitrary nonzero vector inRnand f is differentiable.
Then the proof of Theorem 6.2 goes through without change to give
h→0lim
f(a+hw)− f(a)
h = ∇f(a)·w.
The problem is as follows: Ifw=kvfor some (nonzero) scalark, then
hlim→0
f(a+hw)− f(a)
h = ∇f(a)·w
= ∇f(a)·(kv)
=k(∇f(a)·v)
=k
h→0lim
f(a+hv)− f(a) h
.
That is, the “generalized directional derivative” in the direction ofkvisktimes the derivative in the direction ofv. Butvandkvare parallel vectors, and it is undesirable to have this sort of ambiguity of terminology. So we avoid the trouble by insisting upon using unit vectors only (i.e., by allowingkto be±1 only) when working with directional derivatives.
Gradients and Steepest Ascent
Suppose you are traveling in space near the planet Nilrebo and that one of your spaceship’s instruments measures the external atmospheric pressure on your ship as a function f(x,y,z) of position. Assume, quite reasonably, that this function is differentiable. Then Theorem 6.2 applies and tells us that if you travel from pointa=(a,b,c) in the direction of the (unit) vectoru=ui+vj+wk, the rate of change of pressure is given by
Duf(a)= ∇f(a)·u.
Now, we ask the following: In what direction is the pressure increasing the most?
If θ is the angle between u and the gradient vector ∇f(a), then we have, by Theorem 3.3 of§1.3, that
Duf(a)= ∇f(a) ucosθ = ∇f(a)cosθ,
sinceuis a unit vector. Because−1≤cosθ≤1, we have
−∇f(a) ≤ Duf(a)≤ ∇f(a).
Moreover, cosθ=1 whenθ =0 and cosθ = −1 whenθ=π. Thus, we have established the following:
THEOREM 6.3 The directional derivativeDuf(a) is maximized, with respect to direction, whenupoints in thesamedirection as∇f(a) and is minimized whenu points in theoppositedirection. Furthermore, the maximum and minimum values ofDuf(a) are∇f(a)and−∇f(a), respectively.
EXAMPLE 4 If the pressure function on Nilrebo is f(x,y,z)=5x2+7y4+x2z2atm,
where the origin is located at the center of Nilrebo and distance units are measured in thousands of kilometers, then the rate of change of pressure at (1,−1,2) in the direction ofi+j+kmay be calculated as ∇f(1,−1,2)·u, whereu= (i+j+k)/√
3. (Note that we normalized the vectori+j+kto obtain a unit vector.) Using Theorem 6.2, we compute
Duf(1,−1,2)= ∇f(1,−1,2)·u
=(18i−28j+4k)·i+j+k
√3
= 18−28+4
√3 = −2√
3 atm/Mm.
Additionally, in view of Theorem 6.3, the pressure will increase most rapidly in the direction of∇f(1,−1,2), that is, in the
18i−28j+4k
18i−28j+4k = 9i−14j+2k
√281 direction. Moreover, therateof this increase is
∇f(1,−1,2) =2√
281 atm/Mm. ◆
Theorem 6.3 is stated in a manner that is independent of dimension—that is, so that it applies to functions f:X⊆Rn →Rfor anyn≥2. In the casen=2, there is another geometric interpretation of Theorem 6.3: Suppose you are mountain climbing on the surfacez= f(x,y). Think of the value of f as the height of the mountain above (or below) sea level. If you are equipped with a map and compass (which supply information in thex y-plane only), then if you are at the point on the mountain with x y-coordinates (map coordinates) (a,b), Theorem 6.3 says that you should move in the direction parallel to the gradient∇f(a,b) in order to climb the mountain most rapidly. (See Figure 2.69.) Similarly, you should move in the direction parallel to−∇f(a,b) in order to descend most rapidly. Moreover, the slope of your ascent or descent in these cases is∇f(a,b). Be sure that you understand that∇f(a,b) is a vector inR2 that gives the optimal north–south, east–west direction of travel.
y
y
Map
x
x
z
∇f(a, b)
(a, b) 储∇f(a, b)储
−2 −1 0 1 2
−2
−1 0 1 2
Figure 2.69 Select∇f(a,b)/∇f(a,b)for direction of steepest ascent.
Tangent Planes Revisited
In§2.1, we indicated that not all surfaces can be described by equations of the formz= f(x,y). Indeed, a surface as simple and familiar as the sphere is not the graph of any single function of two variables. Yet the sphere is certainly smooth enough for us to see intuitively that it must have a tangent plane at every point.
(See Figure 2.70.)
Figure 2.70 A sphere and one of its tangent planes.
How can we find the equation of the tangent plane? In the case of the unit spherex2+y2+z2=1, we could proceed as follows: First decide whether the point of tangency is in the top or bottom hemisphere. Then apply equation (4) of
§2.3 to the graph ofz=
1−x2−y2 orz= −
1−x2−y2, as appropriate.
The calculus is tedious but not conceptually difficult. However, the tangent planes to points on the equator are all vertical and so equation (4) of§2.3 does not apply.
(It is possible to modify this approach to accommodate such points, but we will not do so.) In general, given a surface described by an equation of the form F(x,y,z)=c (where c is a constant), it may be entirely impractical to solve for z even as several functions of x and y. Try solving for z in the equation x yz+yex z −x2+yz2 =0 and you’ll see what we mean. We need some other way to get our hands on tangent planes to surfaces described as level sets of functions of three variables.
To get started on our quest, we present the following result, interesting in its own right:
THEOREM 6.4 LetX ⊆Rn be open and f:X →Rbe a function of classC1. Ifx0 is a point on the level set S= {x∈ X| f(x)=c}, then the vector∇f(x0) is perpendicular toS.
PROOF We need to establish the following: Ifvis any vector tangent toSatx0, then∇f(x0) is perpendicular tov(i.e.,∇f(x0)·v=0). By a tangent vector toS atx0, we mean thatvis the velocity vector of a curveC that lies inSand passes throughx0. The situation inR3is pictured in Figure 2.71.
v
x0 C
∇f(x0)
Figure 2.71 The level set surface S= {x| f(x)=c}.
Thus, let C be given parametrically by x(t)=(x1(t),x2(t), . . . ,xn(t)), where a<t <bandx(t0)=x0for some numbert0in (a,b). (Then, ifvis the velocity vector atx0, we must havex(t0)=v. See§3.1 for more about velocity vectors.) SinceCis contained inS, we have
f(x(t))= f(x1(t),x2(t), . . . ,xn(t))=c. Hence,
d
dt [f(x(t))]= d
dt[c]≡0. (4)
On the other hand, the chain rule applied to the composite function f ◦x:
(a,b)→Rtells us d
dt [f(x(t))]= ∇f(x(t))·x(t). Evaluation att0and equation (4) let us conclude that
∇f(x(t0))·x(t0)= ∇f(x0)·v=0,
as desired. ■
Here’s how we can use the result of Theorem 6.4 to find the plane tan- gent to the spherex2+y2+z2 =1 at the point
!−√12,0,√12#
. From§1.5, we know that a plane is determined uniquely from two pieces of information: (i) a point in the plane and (ii) a vector perpendicular to the plane. We are given a point in the plane in the form of the point of tangency
!−√12,0,√12#
. As for a vector normal to the plane, Theorem 6.4 tells us that the gradient of the func- tion f(x,y,z)=x2+y2+z2 that defines the sphere as a level set will do. We have
∇f(x,y,z)=2xi+2yj+2zk, so that
∇f
− 1
√2,0, 1
√2
= −√ 2i+√
2k.
Hence, the equation of the tangent plane is
∇f
− 1
√2,0, 1
√2
·
x+ 1
√2,y−0,z− 1
√2
=0,
−√ 2
x+ 1
√2
+√ 2
z− 1
√2
=0, or
z−x =√ 2.
In general, ifSis a surface inR3defined by an equation of the form f(x,y,z)=c,
then ifx0 ∈ X, the gradient vector∇f(x0) is perpendicular toSand, conse- quently, if nonzero, is a vector normal to the plane tangent to Satx0. Thus, the equation
∇f(x0)·(x−x0)=0 (5) or, equivalently,
fx(x0,y0,z0)(x−x0)+ fy(x0,y0,z0)(y−y0) + fz(x0,y0,z0)(z−z0)=0 (6) is an equation for the tangent plane toSatx0.
Note that formula (5) can be used inRn as well as inR3, in which case it defines the tangent hyperplaneto the hypersurface S⊂Rn defined by f(x1, x2, . . . ,xn)=cat the pointx0 ∈S.
EXAMPLE 5 Consider the surfaceSdefined by the equationx3y−yz2+z5 = 9. We calculate the plane tangent toSat the point (3,−1,2).
To do this, we define f(x,y,z)=x3y−yz2+z5. Then
∇f(3,−1,2)=
3x2yi+(x3−z2)j+(5z4−2yz)k
(3,−1,2)
= −27i+23j+84k
is normal toSat (3,−1,2) by Theorem 6.4. Using formula (6), we see that the tangent plane has equation
−27(x−3)+23(y+1)+84(z−2)=0 or, equivalently,
−27x+23y+84z=64. ◆ EXAMPLE 6 Consider the surface defined by z4 =x2+y2. This surface is the level set (at height 0) of the function
f(x,y,z)=x2+y2−z4. The gradient of f is
∇f(x,y,z)=2xi+2yj−4z3k.
Note that the point (0,0,0) lies on the surface. However,∇f(0,0,0)=0, which makes the gradient vector unusable as a normal vector to a tangent plane. Thus, formula (6) doesn’t apply. What we conclude from this example is that the surface fails to have a tangent plane at the origin, a fact that is easy to believe from the
graph. (See Figure 2.72.) ◆
z
x y
Figure 2.72 The surface of Example 6.
EXAMPLE 7 The equationx2+y2+z2+w2 =4 defines ahypersphereof radius 2 inR4. We use formula (5) to determine the hyperplane tangent to the hypersphere at (−1,1,1,−1).
The hypersphere may be considered to be the level set at height 4 of the function f(x,y,z, w)=x2+y2+z2+w2, so that the gradient vector is
∇f(x,y,z, w)=(2x,2y,2z,2w), so that
∇f(−1,1,1,−1)=(−2,2,2,−2).
Using formula (5), we obtain an equation for the tangent hyperplane as (−2,2,2,−2)·(x+1,y−1,z−1, w+1)=0 or
−2(x+1)+2(y−1)+2(z−1)−2(w+1)=0. Equivalently, we have the equation
x−y−z+w+4=0. ◆
EXAMPLE 8 We determine the plane tangent to the paraboloidz=x2+3y2 at the point (−2,1,7) in two ways: (i) by using formula (4) in§2.3, and (ii) by using our new formula (6).
First, the equation z=x2+3y2 explicitly describes the paraboloid as the graph of the function f(x,y)=x2+3y2, that is, by an equation of the form z= f(x,y). Therefore, formula (4) of§2.3 applies to tell us that the tangent plane at (−2,1,7) has equation
z = f(−2,1)+ fx(−2,1)(x+2)+ fy(−2,1)(y−1) or, equivalently,
z=7−4(x+2)+6(y−1). (7) Second, if we write the equation of the paraboloid as x2+3y2−z =0, then we see that it describes the paraboloid as the level set of height 0 of the three-variable function F(x,y,z)=x2+3y2−z. Hence, formula (6) applies and indicates that an equation for the tangent plane at (−2,1,7) is
Fx(−2,1,7)(x+2)+Fy(−2,1,7)(y−1)+Fz(−2,1,7)(z−7)=0 or
−4(x+2)+6(y−1)−1(z−7)=0. (8) As can be seen, equation (7) agrees with equation (8). ◆ Example 8 may be viewed in a more general context. IfSis the surface inR3 given by the equationz= f(x,y) (where f is differentiable), then formula (4) of
§2.3 tells us that an equation for the plane tangent toSat the point (a,b, f(a,b)) is z= f(a,b)+ fx(a,b)(x−a)+ fy(a,b)(y−b).
At the same time, the equation forSmay be written as f(x,y)−z=0.
Then, if we let F(x,y,z)= f(x,y)−z, we see that Sis the level set of F at height 0. Hence, formula (6) tells us that the tangent plane at (a,b, f(a,b)) is
Fx(a,b, f(a,b))(x−a)+ Fy(a,b, f(a,b))(y−b)
+ Fz(a,b, f(a,b))(z− f(a,b))=0. By construction ofF,
∂F
∂x = ∂f
∂x, ∂F
∂y = ∂f
∂y, ∂F
∂z = −1. Thus, the tangent plane formula becomes
fx(a,b)(x−a)+ fy(a,b)(y−b)−(z− f(a,b))=0.
The last equation for the tangent plane is the same as the one given above by equation (4) of§2.3.
The result shows that equations (5) and (6) extend the formula (4) of§2.3 to the more general setting of level sets.
The Implicit Function and Inverse Function Theorems (optional)
We have previously noted that not all surfaces that are described by equations of the formF(x,y,z)=ccan be described by an equation of the formz= f(x,y).
We close this section with a brief—but theoretically important—digression about when and how the level set{(x,y,z)| F(x,y,z)=c}can also be described as the graph of a function of two variables, that is, as the graph of z= f(x,y).
We also consider the more general question of when we can solve a system of equations for some of the variables in terms of the others.
We begin with an example.
y x
z
(1, 1, −2√3 ) (−2, 2, 6)
Figure 2.73 The two-sheeted hyperboloidz2/4−x2−y2=1.
The point (−2,2,6) lies on the sheet given byz=2
x2+y2+1, and the point (1,1,−2√
3) lies on the sheet given by
z= −2
x2+y2+1.
EXAMPLE 9 Consider the hyperboloid z2/4−x2−y2 =1, which may be described as the level set (at height 1) of the function
F(x,y,z)= z2
4 −x2−y2.
(See Figure 2.73.) This surfacecannotbe described as the graph of an equation of the formz= f(x,y), since particular values forxandygive rise totwovalues forz. Indeed, when we solve forzin terms ofxandy, we find that there are two functional solutions:
z=2
x2+y2+1 and z= −2
x2+y2+1. (9) On the other hand, these two solutions show that, given any particular point (x0,y0,z0) of the hyperboloid, we may solvelocallyfor z in terms ofx andy.
That is, we may identify on which sheet of the hyperboloid the point (x0,y0,z0) lies and then use the appropriate expression in (9) to describe that sheet. ◆ Example 9 prompts us to pose the following question: Given a surfaceS, described as the level set{(x,y,z)| F(x,y,z)=c}, can we always determine at least a portion ofSas the graph of a functionz= f(x,y)? The result that follows,
a special case of what is known as the implicit function theorem, provides relatively mild hypotheses under which we can.
THEOREM 6.5(THE IMPLICIT FUNCTION THEOREM) Let F:X ⊆Rn →Rbe of class C1 and let a be a point of the level set S= {x∈Rn | F(x)=c}. If Fxn(a)=0, then there is a neighborhood U of (a1,a2, . . . ,an−1) in Rn−1, a neighborhood V of an in R, and a function f:U ⊆Rn−1 →V of class C1 such that if (x1,x2, . . . ,xn−1)∈Uandxn ∈VsatisfyF(x1,x2, . . . ,xn)=c(i.e., (x1,x2, . . . ,xn)∈S), thenxn = f(x1,x2, . . . ,xn−1).
The significance of Theorem 6.5 is that it tells us that neara pointa∈S such that∂F/∂xn =0, the level set Sgiven by the equation F(x1, . . . ,xn)=c islocallyalso the graph of a functionxn = f(x1, . . . ,xn−1). In other words, we may solve locally forxn in terms ofx1, . . . ,xn−1, so that Sis, at least locally, a differentiable hypersurface inRn.
EXAMPLE 10 Returning to Example 9, we recall that the hyperboloid is the level set (at height 1) of the functionF(x,y,z)=z2/4−x2−y2. We have
∂F
∂z = z 2.
Note that for any point (x0,y0,z0) in the hyperboloid, we have|z0| ≥2. Hence,
∂Fz(x0,y0,z0)=0. Thus, Theorem 6.5 implies that we may describe a portion of the hyperboloid near any point as the graph of a function of two variables. This
is consistent with what we observed in Example 9. ◆
Of course, there is nothing special about solving for the particular vari- able xn in terms of x1, . . . ,xn−1. Suppose a is a point on the level set S de- termined by the equationF(x)=cand suppose∇F(a)=0. ThenFxi(a)=0 for somei. Hence, we can solve locally nearaforxi as a differentiable function of x1, . . . ,xi−1,xi+1, . . . ,xn. Therefore,S is locally a differentiable hypersurface inRn.
EXAMPLE 11 LetS denote the ellipsoidx2/4+y2/36+z2/9=1. ThenS is the level set (at height 1) of the function
F(x,y,z)= x2 4 + y2
36+z2 9. At the point (√
2,√ 6,√
3), we have
∂F
∂z
(√ 2,√
6,√ 3)
= 2z 9
(√ 2,√
6,√ 3)
= 2√ 3 9 =0. Thus, Smay be realized near (√
2,√ 6,√
3) as the graph of an equation of the form z= f(x,y), namely, z=3
1−x2/4−y2/36. At the point (0,−6,0), however, we see that∂F/∂zvanishes. On the other hand,
∂F
∂y
(0,−6,0)= 2y 36
(0,−6,0)= −1 3 =0.
Consequently, near (0,−6,0), the ellipsoid may be described by solving foryas a function ofx andz, namely, y= −6
1−x2/4−z2/9. ◆
EXAMPLE 12 Consider the set of pointsSdefined by the equationx2z2−y= 0. ThenSis the level set at height 0 of the functionF(x,y,z)=x2z2−y. Note that
∇F(x,y,z)=(2x z2,−1,2x2z).
Since∂F/∂ynever vanishes, we see that we can always solve foryas a function ofxandz. (This is, of course, obvious from the equation.) On the other hand, near points wherex andzare nonzero, both∂F/∂x and∂F/∂zare nonzero. Hence, we can solve for eitherx orzin this case. For example, near (1,1,−1), we have
x = y
z2 and z= − y
x2. ◆
As just mentioned, Theorem 6.5 is actually a special case of a more general result. In Theorem 6.5 we are attempting to solve the equation
F(x1,x2, . . . ,xn)=c
for xn in terms of x1, . . . ,xn−1. In the general case, we have a system of m equations
⎧⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎩
F1(x1, . . . ,xn,y1, . . . ,ym)=c1 F2(x1, . . . ,xn,y1, . . . ,ym)=c2
...
Fm(x1, . . . ,xn,y1, . . . ,ym)=cm
, (10)
and we desire to solve the system fory1, . . . ,ymin terms ofx1, . . . ,xn. Using vec- tor notation, we can also write this system asF(x,y)=c, wherex=(x1, . . . ,xn), y=(y1, . . . ,ym), c=(c1, . . . ,cm), and F1, . . . ,Fm make up the component functions ofF. With this notation, the general result is the following:
THEOREM 6.6(THE IMPLICIT FUNCTION THEOREM,GENERAL CASE) Suppose F:A→Rm is of classC1, where Ais open inRn+m. Let (a,b)=(a1, . . . ,an, b1, . . . ,bm)∈ AsatisfyF(a,b)=c. If the determinant
(a,b)=det
⎡
⎢⎢
⎢⎢
⎣
∂F1
∂y1
(a,b) · · · ∂F1
∂ym
(a,b) ... . .. ...
∂Fm
∂y1 (a,b) · · · ∂Fm
∂ym(a,b)
⎤
⎥⎥
⎥⎥
⎦=0,
then there is a neighborhoodU ofainRn and a unique functionf:U →Rm of classC1 such thatf(a)=bandF(x,f(x))=cfor allx∈U. In other words, we can solve locally foryas a functionf(x).
EXAMPLE 13 We show that, near the point (x1,x2,x3,y1,y2)=(−1,1,1, 2,1), we can solve the system
&
x1y2+x2y1 =1
x12x3y1+x2y23 =3 (11) fory1andy2 in terms ofx1,x2,x3.